In this section, we will explore combinations of H with atoms that have valence 2p electrons, including boron through neon. In later sections we will see what happens when H interacts with the other two atoms from the second row, Li and Be. Here we will start with H and F.

H + F. In the previous section, H⁺, H, and He are all spherical, so there was only one orientation for each pairing. When H intereacts with F, there are more options because F has one singly occupied 2p orbital and two doubly occupied 2p orbitals. The figure shows the possible orientations for H approaching F. We're using the 3D representation that also lets us show the electron spins of the orbitals on H and F.

12.1

In the first row of Figure 12.1 the H atom is shown approaching the singly occupied 2p orbital of F. There are two possibilities for this: the spins of the electrons in the two singly occupied orbitals can be the same (top, left) or different (top, right).

In the bottom row, H approaches one of the doubly occupied orbitals of F. It doesn't matter if the spin of H is up or down, so both diagrams in the bottom row are equivalent.

The plot below shows the potential energy curves for all three cases covered above:

12.2

Only one of the three combinations gives us HF. We see that the correct answer for the first question is (b). Forming HF is a lot like forming H₂ in the previous section: it happens if the spins of the electrons in the two singly occupied orbitals are different. The answer to the second question is also (b). This is the second time we're seen that the interaction between H and a pair of electrons is repulsive.

Figure 12.2 shows the bonding diagrams for the covalent bond in HF. In the 3D representation, we've once again used the convention that when we couple the electrons in singly occupied obitals with spin up (yellow) and spin down (blue) electrons, we have a bond pair that we color gray.

On Figure 12.2 we've put dots to indicate three separations on the potential energy curve for the bound state of HF, at R_e, R_e + 0.4 Å, and R_e + 2.0 Å. Let's look at what happens to the orbitals for H and F at these separations:

12.3

At the longest separation, we see the atomic orbitals we expect to see, the 2p on F and 1s on H. As the atoms come together, both orbitals shift toward the other nucleus and overlap both the orbital and the nucleus. But the shift or polarization is much more pronounced for the H atom orbital. At R_e + 0.4 Å it has already delocalized quite a bit from H onto F. At that separation, the F orbital is barely perturbed toward H. At the minimum separation of R_e, the H orbital has shifted to much that it looks like a distorted 2p orbital. In fact, it resembles the 2p orbital on F quite a bit. The bond in HF is a covalent bond because H and F each contribute an electron and shared them between their nuclei. As in H₂, the bond can form when the electrons have opposite spins because that allows them to overlap between the nuclei. The difference between HF and H₂ is that the sharing in not equal in HF. We call this type of covalent bond a polar covalent bond because the charge is shifted toward one of the atoms (F, in this case).

So far we've only looked at the orbitals that are directly involved in the bond. Now let's look at the other two types of valence orbitals on F, the 2s² pair and the two 2p² pairs:

12.4

The shifts in the orbitals make sense. The F 2s² orbital moves a little bit to the left to get away from the bond pair that forms from the F 1p¹ orbital overlapping the H 1s¹ orbital. The F 2p² hardly changes at all. It moves very, very slightly toward the H nucleus.

Now we will look at animations of all four of these orbitals as linked to the potential energy curve:

12.5

The animations reinforce the comments above and add one new piece of information: the overlap between the two bond orbitals is shown as a bar that increases in length as the bond forms. It ranges from a value close to zero when the atoms are far apart to about 0.8 at R_e.

We'll now look at the overlap as a graph and also show what happens to the dipole moment of HF as the bond forms:

12.6

Both of these plots show an increase in the property as the separation between the nuclei decreases. The dipole moment increases because charge is shifting from H to F.

H + Ne. Let's take a quick look at what happens if H approaches Ne. No matter the orientation, H can only interact with a pair of valence electrons because Ne only has pairs.

You're correct if you choose answer (a). The interaction between H and Ne is repulsive, as shown below.

12.7

H + B–O. All four of the remaining elements in the p-block of the second row of the periodic table have at least one singly occupied orbital. So we would expect it to be possible to to form HX diatomic species from each of them. The figure shows the orientations of the singly occupied orbitals interacting with H as well as the diagrams after covalent bonds have been formed.

12.8

All of these diatomic hydrides can exist, although some of them are quite reactive—the ones with unpaired electrons. Compounds with unpaired electron are called radicals. All of these compounds have polar covalent bonds. We can see how the polarity shifts as we move across the row from B through F:

12.9

In BH and CH, the bond is polarized strongly toward H. As the Z value of the atom increass, the bond becomes more polarized toward the heavy atom and away from H.

This has been a quick overview of how H can interact with the six p-block elements. All of them except Ne can form a polar covalent bond with H.

Let's update the list of bond types we've identified:

The Polar Covalent bond type was added, but we also updated the definitions to call the bonding in H₂ as Nonpolar Covalent.

A terminology note: bonds that involve orbitals whose axes coincide with internuclear axes are known as sigma bonds. In Section 16, we will see another type of bond.

In the next section, we will investigate what happens when H interacts with lithium.